Question

evaluate the limit lim x->5

Answer 1

what is the limit ?

Answer 2

\[\lim_{x \rightarrow 5} \frac{ |x^2-4x-5| }{ x^2-25 }\]

Answer 3

can u show me step by step?

Answer 4

rewrite the top abs value

Answer 5

factor \(\normalsize\color{midnightblue}{ x^2-25=(x-5)(x+5) }\) and factor out the top
\(\normalsize\color{midnightblue}{ │x^2-4x-5│=│(x-5)(x+1)│ }\) and then despite absolute value divide top and bottom by
\(\normalsize\color{midnightblue}{ (x-5) }\)

Answer 6

then plug in 5 instead of x.

Answer 7

but it is 1/2 why is answer dne?

Answer 8

the answer is not 1/2

Answer 9

-1/2?

Answer 10

What do you get after factoring top and bottom and dividing top and bottom by (x-5 )

(without removing the absolute value from the top ) ?

Answer 11

1/2?

Answer 12

it really seems like you are trying to go into calc without knowing even simple algebra.

Answer 13

I am out, jim thompson will help

Answer 14

make a table and you'll find that as x approaches 5 from the left, the limit is different than if you were to approach 5 from the right

So because the left hand limit and the right hand limit are NOT equal, this means the limit as x --> 5 does NOT exist

Answer 15

i think i understand. i think i have to find the values though i cant just say dne?

Answer 16

what if I have \[\lim_{x \rightarrow \infty} \frac{ \cos^5(2x-1) }{ x^3 }\]??

Answer 17

Here's the graph and table I'm getting. I used geogebra to get all this.

Answer 18

Notice the jump discontinuity at x = 5

Answer 19

oh i see that

Answer 20

and notice how the table reflects in this large jump/gap

you should be approaching a single value as you approach 5, but that's not the case

Answer 21

what's the range of cosine?

Answer 22

-1 to 1

Answer 23

so cos^5(x) will also have a range from -1 to 1

because (-1)^5 = -1, (1)^5 = 1

Answer 24

basically, no matter what the argument does, the numerator is going to bounce between -1 and 1

Answer 25

as x ---> infinity, the denominator grows bigger but the numerator is fixed to between -1 and 1

Answer 26

so the whole fraction gets so very very small that it approaches 0

Answer 27

oh right. are you a teacher u teach better than mine!!

Answer 28

I'm studying to become one

I'm glad it's clicking a bit better now

Answer 29

yes. i have a quiz tomorrow. can i ask just one more?

Answer 30

sure

Answer 31

is it same thing for this one? \[\lim_{x \rightarrow 0}x^6\cos^5(\frac{ 1 }{ 5x}- \pi)\]

Answer 32

yeah no matter what, cos is going to bounce around from -1 to 1

as x --> 0, x^6 ---> 0 since x^6 = 0^6 = 0

Answer 33

so we really have 0*cos(x) where x is some number, we really don't need to worry about what its value is

Answer 34

and that whole thing is 0

Answer 35

if they put sine then answer is also same?

Answer 36

thank you i understand tshis a lot better. wow.

Answer 37

yep because they both have the same range

Answer 38

it really applies to any function that has a restricted range